Question

If a metal is irradiated with light of frequency $3\times {10}^{19}$ sec${}^{-1}$, the electron is emitted with a kinetic energy of $6.625\times {10}^{-15}$ J. The threshold frequency of the metal is

• A. $2\times {10}^{19}$ sec${}^{-1}$
• B. $1.25\times {10}^{19}$ sec${}^{-1}$
• C. $6.625\times {10}^{35}$ sec${}^{-1}$
• D. $6.625\times {10}^{19}$ sec${}^{-1}$

Answer 1

The correct option is A $2\times {10}^{19}$ sec${}^{-1}$

According to photoelectric effect

$hv=W+{KE}_{max}⟶\left(1\right)$

given, frequency $v=3\times {10}^{19}{sec}^{-1}$

$K.E=6.625\times {10}^{-15}J$

From $\left(1\right)$, $1.9875\times {10}^{-14}=W+6.625\times {10}^{-15}$

$W=1.325\times {10}^{-14}$

${hv}_0=1.325\times {10}^{-14}$

threshold frequency, $v_0=2\times {10}^{19}{\left(second\right)}^{-1}$